A helicopter has two blades (see figure), each of which has a mass of 270 kg and can be approximated as a thin rod of length 6.7 m. The blades are rotating at an angular speed of 52 rad/s. (a) What is the total moment of inertia of the two blades about the axis of rotation? (a) Determine the rotational kinetic energy of the spinning blades

a. add the moments

b. KE=1/2 Itotal*w^2

To find the total moment of inertia of the two blades about the axis of rotation, we need to calculate the moment of inertia of each blade and then add them together.

The moment of inertia of a thin rod rotating about its perpendicular axis (axis at one end) is given by the formula:

I = (1/3) * m * L^2

where I is the moment of inertia, m is the mass of the rod, and L is the length of the rod.

In this case, the mass of each blade is given as 270 kg and the length of each blade is 6.7 m. So, for each blade:

I1 = (1/3) * m * L^2
= (1/3) * 270 kg * (6.7 m)^2

Now, we need to calculate the total moment of inertia by adding the moments of inertia of both blades:

I_total = I1 + I1
= 2 * I1

To find the rotational kinetic energy of the spinning blades, we can use the formula:

K_rotational = (1/2) * I_total * w^2

where K_rotational is the rotational kinetic energy, I_total is the total moment of inertia, and w is the angular speed.

Substituting the values we have:

K_rotational = (1/2) * (2 * I1) * (52 rad/s)^2

Now you can calculate the total moment of inertia and the rotational kinetic energy using the given values.